Show a certain sequence in Q, with p-adict metric is cauchy

arturo_026
Messages
18
Reaction score
0
I left this following question from my excercises last, hoping that solving the others will give me an insight onto how to proceed. But I still don't have a plan on how to start it:

Consider the sequence s_n = Ʃ (k=0 to n) (t_k * p^k) in Q(rationals) with the p-adic metric (p is prime); where t_n is the sequence 1,2,1,1,2,1,1,1,2,... Show that s_n is Cauchy, but [s_n] (the equivalence class of s_n) cannot be expressed by a rational number.

As far as what I know: I'm familiar with the p-adic metric, and what a cauchy sequence is. I just can't think of how to show that s_n is cauchy.

Thank you very much for any advice and hints.
 
Physics news on Phys.org
Look at |s_n - s_m|. What power of p will divide this?
 
morphism said:
Look at |s_n - s_m|. What power of p will divide this?

Will it be the p-adic absolute value of the partial series from m+1 to n, so that way if I choose N large enough so p^-N is larger or equal to such series (for any n), and p^-N≥ε , then s_n will satisfy the cauchy criterion.
 
Last edited:
I think you have the right idea.
 
morphism said:
I think you have the right idea.

Great! thank you very much. I'll work on cleaning it up.
 
No problem. By the way, in general, a series ##\sum a_n## converges in ##\mathbb Q_p## iff ##a_n \to 0## in the p-adic metric (compare to the case in ##\mathbb R## or ##\mathbb C##!). The proof of this general statement should be similar to the proof you're writing up.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top